Question: $w_1=16[\cos(\dfrac{7\pi}{8})+i\sin(\dfrac{7\pi}{8})]$ $w_2=16[\cos(\dfrac{\pi}{2})+i\sin(\dfrac{\pi}{2})]$ Express the quotient, $\dfrac{w_1}{w_2}$, in polar form. The angle should be given in radians. $\dfrac{w_1}{w_2}= $
Background For any two complex numbers $z_1$ and $z_2$ (whose radii are $r_1$ and $r_2$ and angles are $\theta_1$ and $\theta_2$ ): The radius of $\dfrac{z_1}{z_2}$ is the quotient of the original radii, $\dfrac{r_1}{r_2}$. The angle of $\dfrac{z_1}{z_2}$ is the difference of the original angles, $\theta_1 - \theta_2$. In other words, suppose the polar forms of $z_1$ and $z_2$ are as follows, $z_1 = r_1[\cos(\theta_1) + {i}\sin(\theta_1)]$ $z_2 = r_2[\cos(\theta_2) + {i}\sin(\theta_2)]$, then the polar form of their quotient is: $\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}[\cos(\theta_1 - \theta_2) + {i}\sin(\theta_1 - \theta_2)]$. [How do we get this?] Finding the radius of $\dfrac{w_1}{w_2}$ $w_1=16[\cos(\dfrac{7\pi}{8})+i\sin(\dfrac{7\pi}{8})]$ $w_2=16[\cos(\dfrac{\pi}{2})+i\sin(\dfrac{\pi}{2})]$ Here, $r_1=16$ and $r_2=16$. Therefore, the radius of $\dfrac{w_1}{w_2}$ is $\dfrac{r_1}{r_2}=1$. Finding the angle of $\dfrac{w_1}{w_2}$ $w_1=16[\cos(\dfrac{7\pi}{8})+i\sin(\dfrac{7\pi}{8})]$ $w_2=16[\cos(\dfrac{\pi}{2})+i\sin(\dfrac{\pi}{2})]$ Here, $\theta_1=\dfrac{7\pi}{8}$ and $\theta_2=\dfrac{\pi}{2}$. Therefore, the angle of $\dfrac{w_1}{w_2}$ is $\theta_1-\theta_2=\dfrac{7\pi}{8}-\dfrac{\pi}{2}=\dfrac{3\pi}{8}$ Summary We found that the radius of $\dfrac{w_1}{w_2}$ is $1$ and its angle is $\dfrac{3\pi}{8}$. Therefore, $\dfrac{w_1}{w_2}=1\left(\cos\left(\dfrac{3\pi}{8}\right)+i\sin\left(\dfrac{3\pi}{8}\right)\right)$